Matemática discreta Ejemplos

$\frac{3}{2} x - \frac{1}{4} z = 1$ , $\frac{1}{3} x - \frac{1}{2} z = 2$

Paso 1

Move variables to the left and constant terms to the right.

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Paso 1.1

Simplifica cada término.

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Paso 1.1.1

Combina $\frac{3}{2}$ y $x$ .

$\frac{3 x}{2} - \frac{1}{4} z = 1$

$\frac{1}{3} x - \frac{1}{2} z = 2$

Paso 1.1.2

Combina $z$ y $\frac{1}{4}$ .

$\frac{3 x}{2} - \frac{z}{4} = 1$

$\frac{1}{3} x - \frac{1}{2} z = 2$

$\frac{3 x}{2} - \frac{z}{4} = 1$

$\frac{1}{3} x - \frac{1}{2} z = 2$

Paso 1.2

Reordena los términos.

$\frac{3}{2} x - (\frac{1}{4} z) = 1$

$\frac{1}{3} x - \frac{1}{2} z = 2$

Paso 1.3

Elimina los paréntesis.

$\frac{3}{2} x - \frac{1}{4} z = 1$

$\frac{1}{3} x - \frac{1}{2} z = 2$

Paso 1.4

Simplifica cada término.

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Paso 1.4.1

Combina $\frac{1}{3}$ y $x$ .

$\frac{3}{2} x - \frac{1}{4} z = 1$

$\frac{x}{3} - \frac{1}{2} z = 2$

Paso 1.4.2

Combina $z$ y $\frac{1}{2}$ .

$\frac{3}{2} x - \frac{1}{4} z = 1$

$\frac{x}{3} - \frac{z}{2} = 2$

$\frac{3}{2} x - \frac{1}{4} z = 1$

$\frac{x}{3} - \frac{z}{2} = 2$

Paso 1.5

Reordena los términos.

$\frac{3}{2} x - \frac{1}{4} z = 1$

$\frac{1}{3} x - (\frac{1}{2} z) = 2$

Paso 1.6

Elimina los paréntesis.

$\frac{3}{2} x - \frac{1}{4} z = 1$

$\frac{1}{3} x - \frac{1}{2} z = 2$

$\frac{3}{2} x - \frac{1}{4} z = 1$

$\frac{1}{3} x - \frac{1}{2} z = 2$

Paso 2

Write the system as a matrix.

$[\begin{array}{c} \frac{3}{2} & - \frac{1}{4} & 1 \\ \frac{1}{3} & - \frac{1}{2} & 2 \end{array}]$

Paso 3

Obtén la forma escalonada reducida por filas.

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Paso 3.1

Multiply each element of $R_{1}$ by $\frac{2}{3}$ to make the entry at $1, 1$ a $1$ .

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Paso 3.1.1

Multiply each element of $R_{1}$ by $\frac{2}{3}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{2}{3} \cdot \frac{3}{2} & \frac{2}{3} (- \frac{1}{4}) & \frac{2}{3} \cdot 1 \\ \frac{1}{3} & - \frac{1}{2} & 2 \end{array}]$

Paso 3.1.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{6} & \frac{2}{3} \\ \frac{1}{3} & - \frac{1}{2} & 2 \end{array}]$

Paso 3.2

Perform the row operation $R_{2} = R_{2} - \frac{1}{3} R_{1}$ to make the entry at $2, 1$ a $0$ .

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Paso 3.2.1

Perform the row operation $R_{2} = R_{2} - \frac{1}{3} R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{6} & \frac{2}{3} \\ \frac{1}{3} - \frac{1}{3} \cdot 1 & - \frac{1}{2} - \frac{1}{3} (- \frac{1}{6}) & 2 - \frac{1}{3} \cdot \frac{2}{3} \end{array}]$

Paso 3.2.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{6} & \frac{2}{3} \\ 0 & - \frac{4}{9} & \frac{16}{9} \end{array}]$

Paso 3.3

Multiply each element of $R_{2}$ by $- \frac{9}{4}$ to make the entry at $2, 2$ a $1$ .

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Paso 3.3.1

Multiply each element of $R_{2}$ by $- \frac{9}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{6} & \frac{2}{3} \\ - \frac{9}{4} \cdot 0 & - \frac{9}{4} (- \frac{4}{9}) & - \frac{9}{4} \cdot \frac{16}{9} \end{array}]$

Paso 3.3.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{6} & \frac{2}{3} \\ 0 & 1 & - 4 \end{array}]$

Paso 3.4

Perform the row operation $R_{1} = R_{1} + \frac{1}{6} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Paso 3.4.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{6} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{6} \cdot 0 & - \frac{1}{6} + \frac{1}{6} \cdot 1 & \frac{2}{3} + \frac{1}{6} \cdot - 4 \\ 0 & 1 & - 4 \end{array}]$

Paso 3.4.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & - 4 \end{array}]$

Paso 4

Use the result matrix to declare the final solution to the system of equations.

$x = 0$

$z = - 4$

Paso 5

The solution is the set of ordered pairs that make the system true.

$(0, - 4)$